solve the equations by using
a) the adjoint matrix
b) elementary row operations
i) x+y+z=6
2x-y+z=3
x+4y-z=6
ii) 3p-3q+r=1
-3p+5q-2r=-1
p-2q+r=3
iii)-12a+b+17c=-14
2a-b+3c=5
8a+b-13c=-3
thanks sir/madam
a) Solving equations using the adjoint matrix:
i) To solve the system of equations:
x + y + z = 6 (1)
2x - y + z = 3 (2)
x + 4y - z = 6 (3)
Step 1: Create the coefficient matrix A:
A = [[1, 1, 1],
[2, -1, 1],
[1, 4, -1]]
Step 2: Find the determinant of A, denoted as |A|:
|A| = 1(4(-1) - 1(1)) - 1(2(-1) - 1(1)) + 1(2(1) - 1(4))
= 4 + 3 + 2
= 9
Step 3: Find the matrix of minors:
M = [[3, -2, 3],
[7, -1, -7],
[-5, -1, -1]]
Step 4: Find the matrix of cofactors:
C = [[3, -2, 3],
[-7, 1, -7],
[-5, 1, -1]]
Step 5: Find the adjoint matrix, denoted as adj(A):
adj(A) = C^T (transpose of C)
C^T = [[3, -7, -5],
[-2, 1, 1],
[3, -7, -1]]
So, adj(A) = [[3, -7, -5],
[-2, 1, 1],
[3, -7, -1]]
Step 6: Calculate the inverse matrix A^-1:
A^-1 = adj(A) / |A|
A^-1 = [[3/9, -7/9, -5/9],
[-2/9, 1/9, 1/9],
[3/9, -7/9, -1/9]]
Step 7: Multiply A^-1 with the column vector of constants b = [6, 3, 6]^T:
x = A^-1 * b
x = [[3/9, -7/9, -5/9],
[-2/9, 1/9, 1/9],
[3/9, -7/9, -1/9]] * [6, 3, 6]
Solve this matrix multiplication to get the values of x, y, and z.
ii) Repeat the same steps as above for the second set of equations.
iii) Repeat the same steps as above for the third set of equations.
b) Solving equations using elementary row operations:
i) To solve the system of equations:
x + y + z = 6 (1)
2x - y + z = 3 (2)
x + 4y - z = 6 (3)
Step 1: Write the augmented matrix [A|b]:
A = [[1, 1, 1, 6],
[2, -1, 1, 3],
[1, 4, -1, 6]]
Step 2: Perform elementary row operations to transform the matrix to its row-echelon form or reduced row-echelon form.
Apply row operations to eliminate variables below the main diagonal:
1) R2 = R2 - 2R1
2) R3 = R3 - R1
The new augmented matrix becomes:
A = [[1, 1, 1, 6],
[0, -3, -1, -9],
[0, 3, -2, 0]]
Now, apply row operations to further eliminate variables:
3) R2 = -R2/3
4) R1 = R1 - R2
The new augmented matrix becomes:
A = [[1, 2/3, 2/3, 8],
[0, 1, 1/3, 3],
[0, 3, -2, 0]]
Continue applying row operations until obtaining a form where each leading coefficient is 1 and each column containing a leading coefficient has zeros everywhere else:
5) R3 = R3 - 3R2
The final augmented matrix is:
A = [[1, 0, 1/3, 5],
[0, 1, 1/3, 3],
[0, 0, -3, -9]]
Step 3: Write the system of equations represented by the row-echelon form:
x + (1/3)z = 5 (1)
y + (1/3)z = 3 (2)
-3z = -9 (3)
Step 4: Solve the equations:
From equation (3), z = 3.
Substitute z = 3 into equations (1) and (2):
x + (1/3)(3) = 5
y + (1/3)(3) = 3
Solve for x and y to get the values of x, y, and z.
ii) Repeat the same steps as above for the second set of equations.
iii) Repeat the same steps as above for the third set of equations.